Problem: Simplify $\frac{3^4+3^2}{3^3-3}$  . Express your answer as a common fraction.
Answer: The common factor of 3 in the numerator and the common factor of 3 in the denominator will cancel: \[
\frac{3^4+3^2}{3^3-3}=\frac{3(3^3+3^1)}{3(3^2-1)}=\frac{3^3+3^1}{3^2-1}
\] Now the numerator is $3^3+3=27+3=30$, and the denominator is $3^2-1=9-1=8$.  Therefore, the fraction simplifies to $\dfrac{30}{8}=\boxed{\dfrac{15}{4}}$.